Physics - Semiconductors Question with Solution | TestHub

As Zener is in breakdown region, therefore potential drop across it would be $10 \mathrm{V}$. Now, 

$I_3=\frac{10}{500}=\frac{1}{50} \mathrm{A}$ and 

$I_1=\frac{10}{200}=\frac{1}{20} \mathrm{A}$

Also, $I_2=I_1-I_3$

$\Rightarrow I_2=\left(\frac{1}{20}-\frac{1}{50}\right)=\left(\frac{3}{100}\right) \mathrm{A}=30 \mathrm{mA}$